Question: $f(x, y) = y^2 \sin(x)$ $\dfrac{\partial^2 f}{\partial x^2} = $
Taking a second order partial derivative is like taking a regular second order derivative. We take the partial derivative once, then we take another partial derivative. $\dfrac{\partial^2 f}{\partial x^2} = \dfrac{\partial}{\partial x} \left[ \dfrac{\partial f}{\partial x} \right]$ Let's differentiate! $\begin{aligned} \dfrac{\partial^2 f}{\partial x^2} &= \dfrac{\partial}{\partial x} \left[ \dfrac{\partial}{\partial x} \left[ y^2 \sin(x) \right] \right] \\ \\ &= \dfrac{\partial}{\partial x} \left[ y^2 \cos(x) \right] \\ \\ &= -y^2 \sin(x) \end{aligned}$ Therefore, $\dfrac{\partial^2 f}{\partial x^2} = -y^2 \sin(x)$.